Katie asks further about the case where we have a subset $A$ of $\mathbb{Z}/n$ and asks for $\sum_{k \in A} \zeta^k$ to have integer absolute value. She writes that, for $n$ prime, the only solutions should be the trivial ones $|A| =0$, $1$, $p-1$ or $p$. The point of this note is that she is correct, and that there probably isn't a good description like this for $n$ not prime. This answer is built on Noam Elkies's very helpful comment above.
Let $p$ be prime and let $A \subseteq \mathbb{Z}/p$. Let $a=|A|$. Let $z=\sum_{k \in A} \zeta^k$. Let $b_k = \# \{ (i,j) \in A^2 : i-j \equiv k \bmod p \}$ So $$z \bar{z} = a + \sum_{k=1}^{p-1} b_k \zeta^k$$ Since the minimal polynomial of $\zeta$ is $1+\zeta+\zeta^2+\cdots+\zeta^{p-1}$, the only way that $z \bar{z}$ can be an integer is if all the $b_k$ are equal to some common value, say $b$. In this case, $z \bar{z} = a-b$. Furthermore, we want $\sqrt{z \bar{z}}$ to be an integer, say $n$. So $a=n^2+b$ for some nonnegative integer $n$.
Now, we must have $a(a-1) = b(p-1)$ since $\sum_{k=1}^{p-1} b_k$ is clearly $a(a-1)$. So $(n^2+b)(n^2+b-1) = b(p-1)$. If $b=0$ then $|A|=0$ or $1$. If not, we can divide by $b$ to write: $$p=\frac{(n^2+b)(n^2+b-1)}{b}+1 = \frac{(n^2+n+b)(n^2-n+b)}{b}.$$ (That factorization came out of nowhere, as far as I'm concerned.) Since $p$ is prime, that means that at least one of $n^2+n+b$ and $n^2-n+b$ is $\leq b$. But $n^2 \pm n \geq 0$ for nonnegative integer $n$. So $n=0$ or $1$. This gives $p=b$ and $p=b+2$, and then $a=p$ and $a=p-1$ respectively.
Now, if $n$ is not prime, then there are lots of sets $A$ such that all the $b_k$ are equal. The special case $b=1$ is called a perfect difference set, and they are pretty plentiful. For example, $(0,5,6,9,19)$ is a perfect difference set modulo $21$, giving $$|1+\zeta_{21}^5+\zeta_{21}^6+\zeta_{21}^9+\zeta_{21}^{19}| = \sqrt{5-1} = 2.$$ According to the mathworld link above, there are perfect difference sets modulo $q^2+q+1$ for every perfect power $q$; taking $q=p^2$ we deduce that we can always find $A$ in $\mathbb{Z}/(p^4+p^2+1)$ with absolute value $p$.
Based on skimming the results of a quick google search, my impression is that there are lots of methods known for constructing perfect difference sets, but no classification. Probably someone has studied the case $b>1$, but I didn't see it.
But even if there were a classification, that wouldn't be a complete answer. Because $1+\zeta+\cdots + \zeta^{n-1}$ is not the minimal polynomial of $\zeta$ for $n$ composite. And I don't see how to control the other solutions that might come up because of this.